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any math people... (pg. 3)
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| Cloudburst |
| I think he should adjust his shield harmonics to compensate for any future math problems. However that could cause temporal anomalies in the flux capacitor. Caution is advised. |
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| Krypton |
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| Trancealot |
| quote: | Originally posted by Zild
Does the problem instruct you to factor the equation? Or is it up to you to figure out which method?
If you're supposed to factor then you copied the problem incorrectly or there is a typo somewhere because you can't factor that one. |
he says (you may want to check the results using software as matlab)...
The actual question is find the eigenvalues and eigenvectors or matrix A;
A=[3 -1 1;-1 8 -2;1 -2 5]
doing this gives you cubic polynomial in the beginning. I just wanted to know if it was possible to breakdown and get nice looking roots even with weird fractions whic are those results shown for x1,x2,x3 because I hate getting x^3+... on a test without a program or stuff in front of me!! |
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| Zild |
| quote: | Originally posted by Trancealot
he says (you may want to check the results using software as matlab)...
The actual question is find the eigenvalues and eigenvectors or matrix A;
A=[3 -1 1;-1 8 -2;1 -2 5]
doing this gives you cubic polynomial in the beginning. I just wanted to know if it was possible to breakdown and get nice looking roots even with weird fractions whic are those results shown for x1,x2,x3 because I hate getting x^3+... on a test without a program or stuff in front of me!! |
Ahh then you didn't copy it wrong, and I don't know any way to factor it easily. |
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| Axer |
Find x and replace it with it's new value.
Since the equation equals zero, therefore you can probably from other equations with it and solve is simultaneously, don't know...just guessing. |
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| diesel_tron3000 |
| quote: | Originally posted by Trancealot
he says (you may want to check the results using software as matlab)...
The actual question is find the eigenvalues and eigenvectors or matrix A;
A=[3 -1 1;-1 8 -2;1 -2 5]
doing this gives you cubic polynomial in the beginning. I just wanted to know if it was possible to breakdown and get nice looking roots even with weird fractions whic are those results shown for x1,x2,x3 because I hate getting x^3+... on a test without a program or stuff in front of me!! |
the factors you get from the program you use are the eigenvalues. you need these to find the eigenvectors.
eigenvalues:
9.288
2.579
4.1331
for the eigenvectors of which there are three, use the eigenvalue matrix with the appropriate substitutions for x1, x2, x3, and reduce them down to one equation.
eigenvalue matrix:
|(3-x) -1 1|
|-1 (8-x) -2|
|1 -2 (5-x)|
kind of looks like a ty matrix but you can see how it lines up i guess
x=9.288
vector=(-6.288, -4.288, -5.288)
x=4.1331
vector=(-1.1331, 0.87, -0.13)
x=2.579
vector=(0.43, 2.43, 1.43)
if you want work to show, i guess you should diagonalize it and do all that other e to prove it's rotation or stretching
are there vB math codes? |
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| Leif |
| polynomial long division is indeed the way to go. Easier than regular long division :p |
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| Trancealot |
| quote: | Originally posted by diesel_tron3000
the factors you get from the program you use are the eigenvalues. you need these to find the eigenvectors.
eigenvalues:
9.288
2.579
4.1331
for the eigenvectors of which there are three, use the eigenvalue matrix with the appropriate substitutions for x1, x2, x3, and reduce them down to one equation.
eigenvalue matrix:
|(3-x) -1 1|
|-1 (8-x) -2|
|1 -2 (5-x)|
kind of looks like a ty matrix but you can see how it lines up i guess
x=9.288
vector=(-6.288, -4.288, -5.288)
x=4.1331
vector=(-1.1331, 0.87, -0.13)
x=2.579
vector=(0.43, 2.43, 1.43)
if you want work to show, i guess you should diagonalize it and do all that other e to prove it's rotation or stretching
are there vB math codes? |
1. this is the easy part you showed(having answer) and finishing it
2. The hard part is what I was asking from the get go with the x^3... Seems from people's suggestion is can not be factored to get the 3 roots excpet with the use of a semi-crazy method shown in a previous post in this thread.
3. Good attempt. I can't complain from the work you put in. :) |
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| spacechica |
Hi Trancealot,
Did you find what you were looking for?
Best,
spacechica |
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| idoru |
| This thread is making my ing head explode. :mad: |
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| SuspicionVandit |
| quote: | Originally posted by Trancealot
Thanks if anyone can get me in the right direction... |
I'm just kidding :) |
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